# What is int ( 2 + x ) / sqrt ( 4 - 2x - x^2 dx?

$\int \frac{2 + x}{\sqrt{4 - 2 x - {x}^{2}}} \mathrm{dx} = \arcsin \left(\frac{x + 1}{\sqrt{5}}\right) - \sqrt{4 - 2 x - {x}^{2}} + C$

#### Explanation:

Start from the given

$\int \frac{2 + x}{\sqrt{4 - 2 x - {x}^{2}}} \mathrm{dx}$

$4 - 2 x - {x}^{2} = - \left({x}^{2} + 2 x - 4\right) = - \left({x}^{2} + 2 x + 1 - 1 - 4\right)$

and

$4 - 2 x - {x}^{2} = - \left({\left(x + 1\right)}^{2} - 5\right) = 5 - {\left(x + 1\right)}^{2}$

then

$\int \frac{2 + x}{\sqrt{4 - 2 x - {x}^{2}}} \mathrm{dx} = \int \frac{2 + x}{\sqrt{5 - {\left(x + 1\right)}^{2}}} \mathrm{dx}$

The Trigonometric Substitution

Let $x + 1 = \sqrt{5} \cdot \sin \theta$
and $x = \sqrt{5} \cdot \sin \theta - 1$
and $\mathrm{dx} = \sqrt{5} \cdot \cos d \theta$

Let's do the substitution

$\int \frac{2 + x}{\sqrt{5 - {\left(x + 1\right)}^{2}}} \mathrm{dx} =$
$\int \frac{\left(\sqrt{5} \sin \theta + 1\right) \cdot \left(\sqrt{5} \cdot \cos \theta\right) d \theta}{\sqrt{5 - {\left(\sqrt{5} \cdot \sin \theta\right)}^{2}}}$

$\int \frac{\left(\sqrt{5} \sin \theta + 1\right) \cdot \left(\sqrt{5} \cdot \cos \theta\right) d \theta}{\sqrt{5 - 5 \cdot {\sin}^{2} \theta}}$

continue simplification by trigonometric identities

$\int \frac{\left(\sqrt{5} \sin \theta + 1\right) \cdot \left(\sqrt{5} \cdot \cos \theta\right) d \theta}{\sqrt{5} \cdot \sqrt{1 - {\sin}^{2} \theta}}$

$\int \frac{\left(\sqrt{5} \sin \theta + 1\right) \cdot \left(\sqrt{5} \cdot \cos \theta\right) d \theta}{\sqrt{5} \cdot \sqrt{{\cos}^{2} \theta}}$

$\int \frac{\left(\sqrt{5} \sin \theta + 1\right) \cdot \left(\sqrt{5} \cdot \cos \theta\right) d \theta}{\sqrt{5} \cdot \cos \theta}$

$\int \frac{\left(\sqrt{5} \sin \theta + 1\right) \cdot \left(\sqrt{5} \cdot \cos \theta\right) d \theta}{\sqrt{5} \cdot \cos \theta}$

and

$\int \frac{2 + x}{\sqrt{5 - {\left(x + 1\right)}^{2}}} \mathrm{dx} = \int \left(\sqrt{5} \sin \theta + 1\right) d \theta$

$\int \frac{2 + x}{\sqrt{5 - {\left(x + 1\right)}^{2}}} \mathrm{dx} = \int \left(1 + \sqrt{5} \sin \theta\right) d \theta$

$\int \frac{2 + x}{\sqrt{5 - {\left(x + 1\right)}^{2}}} \mathrm{dx} = \theta - \sqrt{5} \cdot \cos \theta + C$

Now, time to imagine your right triangle with
angle $\theta$
Let $x + 1$ the Opposite side to angle $\theta$
Let $\sqrt{5}$ the Hypotenuse
Let $\sqrt{4 - 2 x - {x}^{2}}$ the Adjacent side to angle $\theta$

Return the variables

$\int \frac{2 + x}{\sqrt{5 - {\left(x + 1\right)}^{2}}} \mathrm{dx} = \theta - \sqrt{5} \cdot \cos \theta + C$

$\int \frac{2 + x}{\sqrt{5 - {\left(x + 1\right)}^{2}}} \mathrm{dx} = \arcsin \left(\frac{x + 1}{\sqrt{5}}\right) - \sqrt{4 - 2 x - {x}^{2}} + C$

I hope the explanation is useful....God bless...